Precalculus Examples

$\log_{8} (x + 6) = 1 - \log_{8} (x + 4)$

Step 1

Move all the terms containing a logarithm to the left side of the equation.

$\log_{8} (x + 6) + \log_{8} (x + 4) = 1$

Step 2

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

$\log_{8} ((x + 6) (x + 4)) = 1$

Step 3

Expand $(x + 6) (x + 4)$ using the FOIL Method.

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Step 3.1

Apply the distributive property.

$\log_{8} (x (x + 4) + 6 (x + 4)) = 1$

Step 3.2

Apply the distributive property.

$\log_{8} (x \cdot x + x \cdot 4 + 6 (x + 4)) = 1$

Step 3.3

Apply the distributive property.

$\log_{8} (x \cdot x + x \cdot 4 + 6 x + 6 \cdot 4) = 1$

Step 4

Simplify and combine like terms.

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Step 4.1

Simplify each term.

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Step 4.1.1

Multiply $x$ by $x$ .

$\log_{8} (x^{2} + x \cdot 4 + 6 x + 6 \cdot 4) = 1$

Step 4.1.2

Move $4$ to the left of $x$ .

$\log_{8} (x^{2} + 4 \cdot x + 6 x + 6 \cdot 4) = 1$

Step 4.1.3

Multiply $6$ by $4$ .

$\log_{8} (x^{2} + 4 x + 6 x + 24) = 1$

Step 4.2

Add $4 x$ and $6 x$ .

$\log_{8} (x^{2} + 10 x + 24) = 1$

Step 5

Rewrite $\log_{8} (x^{2} + 10 x + 24) = 1$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$8^{1} = x^{2} + 10 x + 24$

Step 6

Solve for $x$ .

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Step 6.1

Rewrite the equation as $x^{2} + 10 x + 24 = 8^{1}$ .

$x^{2} + 10 x + 24 = 8$

Step 6.2

Subtract $8$ from both sides of the equation.

$x^{2} + 10 x + 24 - 8 = 0$

Step 6.3

Subtract $8$ from $24$ .

$x^{2} + 10 x + 16 = 0$

Step 6.4

Factor $x^{2} + 10 x + 16$ using the AC method.

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Step 6.4.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $16$ and whose sum is $10$ .

$2, 8$

Step 6.4.2

Write the factored form using these integers.

$(x + 2) (x + 8) = 0$

Step 6.5

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x + 2 = 0$

$x + 8 = 0$

Step 6.6

Set $x + 2$ equal to $0$ and solve for $x$ .

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Step 6.6.1

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Step 6.6.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 6.7

Set $x + 8$ equal to $0$ and solve for $x$ .

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Step 6.7.1

Set $x + 8$ equal to $0$ .

$x + 8 = 0$

Step 6.7.2

Subtract $8$ from both sides of the equation.

$x = - 8$

Step 6.8

The final solution is all the values that make $(x + 2) (x + 8) = 0$ true.

$x = - 2, - 8$

Step 7

Exclude the solutions that do not make $\log_{8} (x + 6) = 1 - \log_{8} (x + 4)$ true.

$x = - 2$